Probability Proof

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• Mar 7th 2011, 01:34 PM
Mcoolta
Probability Proof
I am given two events, A and B, and a new event C occurs, if and only if exactly one of A or B occurs. I need to prove;

$Pr(C) = Pr(A) + Pr(B) - 2Pr(A\cap B)$

So from the question, i got:

$Pr(C) = Pr (A \cap \bar {B}) \cup Pr(\bar{A} \cap B)$

Is that correct and a good starting point? Or am i missing something out?
• Mar 7th 2011, 02:23 PM
Plato
From the second equation you can use:
$P(A\cap\overline{B})=P(A)-P(A\cap B)$.
• Mar 8th 2011, 09:03 AM
Mcoolta
So i get:

$(P(A)-P(A\cap B))\cup(P(B)-P(A \cap B))$

Is that correct?

How do i go about getting rid of the union in the middle?
• Mar 8th 2011, 09:20 AM
Plato
Quote:

Originally Posted by Mcoolta
So i get:

$(P(A)-P(A\cap B))\cup(P(B)-P(A \cap B))$

How do i go about getting rid of the union in the middle?

Don't you get
$(P(A)-P(A\cap B))+(P(B)-P(A \cap B))~?$
• Mar 8th 2011, 09:49 AM
Mcoolta
Im confused to where the union has gone, from my original post?

Thanks for your help so far!
• Mar 8th 2011, 09:56 AM
Plato
Are you not proving that $P([A\cap\overline{B}]\cup[B\cap\overline{A}])=P(A)+P(B)-2P(A\cap B)~?$
That is what I thought you were doing. Is that not right?
• Mar 8th 2011, 01:35 PM
Mcoolta
Yeah that is what im proving, im just confused from where the middle '+' has came from in the 4th post down, instead of a $\cup$:

Why is it:

$P[(A)-(A \cap B)] '+' [(B)-(A \cap B)]$

Instead of

$P[(A)-(A \cap B)] '\cup' [(B)-(A \cap B)]$

Sorry if that is unclear, Thanks
• Mar 8th 2011, 01:48 PM
Plato
If you know that $K\cap J=\emptyset$ then you know that $P(K\cup J)=P(K)+P(J)$. Right?

Now $(A\cap\overline{B})\cap(B\cap\overline{A})=\emptys et.$

So the probability of that union is the sum of the probabilities.